Hydrochloric acid (Mr = 36.5) reacts with calcium carbonate (Mr = 100) to make calcium chloride (Mr = 75.5), water (Mr = 18) and carbon dioxide (Mr = 44), according to the following equation:CaCO3 + 2HCl → CaCl2 + H2O + CO2At STP, 5 g of calcium carbonate reacts and produces 330 cm3 of carbon dioxide. What is the % yield of carbon dioxide in this reaction?Molar volume of a gas at STP = 22.7 dm3 = 22700 cm370.9%0.29%33.0%29.1%
Question
Hydrochloric acid (Mr = 36.5) reacts with calcium carbonate (Mr = 100) to make calcium chloride (Mr = 75.5), water (Mr = 18) and carbon dioxide (Mr = 44), according to the following equation:CaCO3 + 2HCl → CaCl2 + H2O + CO2At STP, 5 g of calcium carbonate reacts and produces 330 cm3 of carbon dioxide. What is the % yield of carbon dioxide in this reaction?Molar volume of a gas at STP = 22.7 dm3 = 22700 cm370.9%0.29%33.0%29.1%
Solution
To solve this problem, we need to follow these steps:
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Calculate the number of moles of calcium carbonate (CaCO3) used: The molar mass of CaCO3 is 100 g/mol. So, 5 g of CaCO3 is 5/100 = 0.05 mol.
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According to the balanced chemical equation, one mole of CaCO3 produces one mole of CO2. So, 0.05 mol of CaCO3 should theoretically produce 0.05 mol of CO2.
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Calculate the theoretical volume of CO2 produced: At STP, one mole of any gas occupies 22.7 dm3 (or 22700 cm3). So, 0.05 mol of CO2 should occupy 0.05 * 22700 = 1135 cm3.
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Calculate the percentage yield: The actual volume of CO2 produced is given as 330 cm3. So, the percentage yield = (actual yield/theoretical yield) * 100 = (330/1135) * 100 = 29.1%.
So, the % yield of carbon dioxide in this reaction is 29.1%.
Similar Questions
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